Let's say there are 2 basis spanning a subspace. One of the basis is a scalar multiple of the other basis?

Question

So, can the vector that is a scalar multiple of one of the basis out of the 2 basis vectors spanning the subspace?

I dont know how to paraphrase my question. In case, you dont get my question,

I give an example here: let's say the subspace is spanned by {(1,2,3),(2,4,6)} Can (3,6,9) be considered a basis? I don't think so, because a basis by definition means it must not be linearly dependent of the 2 vectors.

Answer 1

Since $(3,6,9)$ is a vector, it cannot be a basis. But, yes, $\bigl\{(3,6,9)\bigr\}$ is a basis of the subspace of $\mathbb R^3$ spanned by $\bigl\{(1,2,3),(2,4,6)\bigr\}$. Note that this subspace is simply $\mathbb R(1,2,3)$ and that $(3,6,9)=3(1,2,3)$.

Answer 2

Let $V$ be a vector space over a field $K$ and let $u,v \in V, k \in K$ and $v=ku.$

If $u \ne 0$ then the vector subspace spanned by $\{u,v\} $ is the the one-dimensional subspace $\{tu: t \in K\}$.

If $u \ne 0$ and $k \ne 0$, then $\{tu: t \in K\}=\{sv: s \in K\}.$